'Mystery Matrix' printed from http://nrich.maths.org/

We had many good solutions sent in explaining some of the ways the answers were found. First from Romy from St Augustine's Catholic Primary School in Solihul we had this:l

Here is how I got to my solution:

First, I had a look at all the numbers that I had already been given. From these, I noticed the number 49. This number is special because it is a square number which means it is the same number times by itself. Because of this, I knew that the two numbers that created it were 7 and 7. I entered this into my table.

Next, I used the 7s I had found and I found that 42 went into 7 6 times. Because I knew this, I could figure out that the number in the bottom left was a 6. The numbers that were already there were really helpful, it made it all be much easier.

After that, I looked at the number 32. I found 1, 32, 2, 16, 4 and 8 as its factors. I chose 4 and 8 because 4 was also divisible by 40, which was in the same row. 40 was a number that was already given to me and, by dividing it by 4, I soon found out that the number above it would be 10. That helped me find out the other answers by doing 10 x 7 and 10 x 6.

At this point, I had quite a few numbers in my grid, which helped me find the rest. This made it easier. I could then do 7 x 8 and 6 x 8. I carried on this process until I could not find any more numbers. When this happened I looked at other numbers I had already been given. I found the
factors of these numbers and tested them out until I found a solution that worked.

Finally, when I had found all the numbers I needed to multiply by, I just had to work out the answers to them and write them in the chart.

I liked this challenge very much and I hope others will use my solution. I had to use my knowledge from the past of things such as:

Division helped me to check and see if the numbers I had written were correct and divisible by the numbers at the top and the side. I really enjoyed doing this challenge and I am pleased with my result.

This is how we did it:
First look at the numbers in red and find a number that can only have one solution. 49 can only be made by 7x7 so that is the first number to be completed.
Now you can enter 6 in the bottom row of the column as 6x7 = 42.
Next if you look at the fourth row both numbers are only in the 3x table. So now we can enter 5 in the 2nd column as 5x3 = 15 also 9 in the last column (9x3 =27).
As we have to use 7 twice, we can't use any other number twice so we know that 10 must be at the top of the 4th column (not 5) and 4 on the first row.
This means 8 must be on the top of the 1st column (8x4=32). 22 and 24 are in the 2 x table so the 3rd column must be 2 which means 11 must be in the 3rd row and 12 in the 5th row.

Children from St Nicholas School sent in this picture ;

They then wrote;

Across the top we worked out: 8, 5, 2, 10, 7, 9
Down the side we worked out 4, 7, 11, 3, 12, 6

We used a combination of trial and error, and knowledge of factors of numbers.
We used knowledge of times tables too.

Thank you all for sending in your solutions, its good for both you and us to read of your explanations.